VOL. 79 1999 NO. 1
ON THE METRIC THEORY OF CONTINUED FRACTIONS
BY
JO¨ EL R I V A T (LYON)
Introduction. For any positive integer n we denote by P (n) the Lebe- sgue measure of the set of irrational numbers x ∈ (0, 1) whose closest ratio- nal approximation with denominator ≤ n is a convergent of the continued fraction expansion of x.
The question of the behaviour of P (n) was asked by M. Del´eglise to A. Schinzel. Recently I. Aliev, S. Kanemitsu and A. Schinzel [1] proved that
P (n) = 1 2 + 6
π 2 (log 2) 2 + O 1 n
.
In this article we shall improve this result to the following Theorem . There exists c > 0 such that
(1) P (n) = 1 2 + 6
π 2 (log 2) 2 + O 1 n exp
−c (log n) 3/5 (log log n) 1/5
. Under the Riemann hypothesis we have
(2) P (n) = 1
2 + 6
π 2 (log 2) 2 + O(n −4/3+ε ).
Remark . I. Aliev, S. Kanemitsu and A. Schinzel [1] also note that the main term, but not the error term, can be derived from Theorem 1.3 of P. Kargaev and A. Zhigljavsky [2].
Classical results. We denote by ⌊x⌋ the greatest integer not exceeding x and write ψ(x) = x − ⌊x⌋ − 1/2.
Lemma 1. Let f be a function with a continuous derivative in the interval [a, b]. Then
X
a<n≤b
f (n) =
b
\
a
f (x) dx + ψ(a)f (a) − ψ(b)f (b) +
b
\
a
ψ(x)f ′ (x) dx.
P r o o f. See for example Titchmarsh [4], formula 2.1.2, page 13.
1991 Mathematics Subject Classification: Primary 11K50.
This work was started while the author was visiting the Polish Academy of Sciences in Warsaw.
[9]
Applying this lemma and writing φ(x) = ψ(x)/x, log + x = max(log x, 1) we obtain
Lemma 2. For arbitrary positive numbers a < b we have X
a<n≤b
1
n = log b − log a + φ(a) − φ(b) + O 1 a 2
,
X
a<n≤b
log n
n = (log b) 2 − (log a) 2
2 + φ(a) log a − φ(b) log b + O log + a a 2
, X
a<n≤b
1 n 2 = 1
a − 1
b + O 1 a 2
.
Corollary 1. For any positive number x, we have X
x/2<k≤x
1
k = log 2 + φ x 2
− φ(x) + O 1 x 2
,
X
x/2<k≤x
log k
k = log x 2
log 2 + (log 2) 2 2 + φ x
2
log x
2
− φ(x) log x + O log + x x 2
, X
x/2<k≤x
1 k 2 = 1
x + O 1 x 2
.
Lemma 3. There exists c > 0 such that for any x ≥ 1 we have X
1≤d≤x
µ(d) d = O
exp
−c (log x) 3/5 (log log x) 1/5
,
X
1≤d≤x
µ(d) d 2 = 6
π 2 + O 1 x exp
−c (log x) 3/5 (log log x) 1/5
. Under the Riemann hypothesis, for any x ≥ 1 we have
X
1≤d≤x
µ(d)
d = O(x −1/2+ε ), X
1≤d≤x
µ(d) d 2 = 6
π 2 + O(x −3/2+ε ).
P r o o f. By partial summation, for any 1 ≤ x ≤ y we have X
x<d≤y
µ(d) d = 1
y X
x<d≤y
µ(d) +
y
\
x
X
x<d≤t
µ(d) dt t 2 , X
x<d≤y
µ(d) d 2 = 1
y 2 X
x<d≤y
µ(d) + 2
y
\
x
X
x<d≤t
µ(d) dt
t 3 .
By Satz 3 of A. Walfisz [5], page 191, there exists c ′ > 0 such that X
1≤d≤x
µ(d) = O
x exp
−c ′ (log x) 3/5 (log log x) 1/5
. Writing
δ(t) = exp
−c ′ (log t) 3/5 (log log t) 1/5
we have P
x<d≤t µ(d) ≪ tδ(t), hence X
x<d≤y
µ(d)
d ≪ δ(y) +
y
\
x
δ(t) dt t
≪ δ(y) + δ(x)(log x) 2
y
\
x
dt
t(log t) 2 ≪ δ(x) log x, X
x<d≤y
µ(d)
d 2 ≪ δ(y) y +
y
\
x
δ(t) dt
t 2 ≪ δ(y)
y + δ(x)
y
\